The equation of a circle $C$ is $x^2+y^2-6x-6y+9 = 0$. What is its center $(h, k)$ and its radius $r$ ?
Explanation: To find the equation in standard form, complete the square. $(x^2-6x) + (y^2-6y) = -9$ $(x^2-6x+9) + (y^2-6y+9) = -9 + 9 + 9$ $(x-3)^{2} + (y-3)^{2} = 9 = 3^2$ Thus, $(h, k) = (3, 3)$ and $r = 3$.